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Here is how my textbook proves that dissolution of solid in liquid is dynamic in nature. It considers the example of a saturated solution of sugar. It says that,

Though the solution is in equilibrium, the process of dissolution of sugar and crystallization of sugar takes place continuously and also at the same rate. This can be confirmed with the help of radioactive sugar. If we drop some radioactive sugar into saturated solution of non-radioactive sugar, then after some time radioactivity is observed both in the solution and in the solid sugar. Initially there were no radioactive sugar molecules in the solution but due to dynamic nature of equilibrium, there is exchange between the radioactive and non-radioactive sugar molecules between the two phases.

It also proves the dynamic nature of chemical equilibrium in a similar way.

From this proof, my argument is that dynamic nature of equilibrium is proved only by adding another substance. This disturbs the equilibrium. It is possible that the saturated solution was in static equilibrium but addition of this substance has led to the dynamic nature.
I do not find it logical to prove that it is dynamic by the process of disturbing it.

Is this proof invalid or am I going wrong in my arguments?

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  • $\begingroup$ For my clarification, why do you say that another substance is being added? To me, the description of the experiment says sugar (with radioactive carbon) is added to a saturated solution; sugar is being added to a sugar solution. $\endgroup$ – J. Ari Apr 25 '17 at 19:19
  • $\begingroup$ In that statement, I was referring to how it is proved generally (in chemical equilibrium as well). When I said another substance is added, I meant that - always something (same or different) is introduced from outside. And still isn't sugar with radioactive carbon different from that of normal carbon? $\endgroup$ – DoubtExpert Apr 26 '17 at 1:58
  • $\begingroup$ The importance is in how it is different. In this case, the different isotope of carbon will not be so different as to perceptibly affect the dissolution process. $\endgroup$ – J. Ari Apr 26 '17 at 14:34
  • $\begingroup$ One would have the same effect by having two identical solutions (saturated, some solid sugar at the bottom), except that one is of radioactive sugar, with a removable wall in between. Some time after the removal, one would find radioactive sugar at the bottom of the container that initially only contained non-radioactive sugar. $\endgroup$ – TAR86 Apr 26 '17 at 19:49
  • $\begingroup$ @TAR86 I am not sure whether on removal of the wall, the combined system (of two equilibria) is immediately in equilibrium. If this is false, then according to your example isn't this a disturbance? If the systems were assumed to be in static equilibrium, then due to this disturbance, it can be argued that the reactions would activate and this would lead radioactive sugar to be part of other container. $\endgroup$ – DoubtExpert May 7 '17 at 16:44
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You are technically correct in saying that radioactive sugar is another species in that it is possible to distinguish it from 'normal' sugar, although chemically identical. But think about it this away. Some of the added radioactive sugar solution (which could be micromolar or less) finds its way into the solid sugar at the bottom of the flask, but some remains in solution so that (dynamic) equilibrium is reached with respect to the radioactive sugar. You could do it the other way round, add solid radioactive sugar and it will appear in the solution. What is so special about the radioactive sugar that it can do this and 'normal' sugar cannot? Well, there is nothing special so it is entirely reasonable to suppose that the normal sugar does this also. By inference, other molecules also do this as we know of no experimental reason why sugar is peculiar among all the other types of molecules known.

If you are still not convinced an experiment could be done with a coloured salt such as copper sulfate, so only one species. Make a saturated solution in which there will be solid in equilibrium with water, which will be blue due to the copper ions. Next, filter off the water and retain the solid and (gently) add more water, initially this will be clear but becomes increasingly blue as the sulfate dissolves. You will see equilibrium being established again as the sulfate dissolves, and provided there is some solid left equilibrium solid to solution will be achieved.

There is an experimental technique called temperature jump that is used specifically to study the time it takes to 'relax' back to equilibria and so obtain rate constants for the process. In this method, a short duration laser pulse is used to rapidly heat a solution (typically by $\ll 1$ C) and change its equilibrium a little. The return to equilibrium is followed by optical/spectroscopic methods so nothing has to be taken from the solution. Many reactions have been studies this way e.g. $\ce{H3O+ + OH^- <=> 2H2O}$.

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This experiment doesn't disturb the equilibrium and is a good way to demonstrate its dynamic nature

The key point in this experiment is that it doesn't disturb the equilibrium. Radioactive sugar is–for all practical reasons that matter to the experiment–chemically identical to the sugar in the solution. So, from a chemical point of view, the equilibrium is not disturbed at all by the addition of radioactive solid sugar.

If the equilibrium were static (meaning that no molecules were exchanged between the solid and the liquid once the solution is fully saturated) then there would be no exchange between the solid and the liquid. If the equilibrium is dynamic, then there will be. If we tried to test this with identical sugar molecules then we would be stuck as there is no way to tell whether the molecules exchanged between the solid state and the solution (because all sugar molecules are identical). Using radioactive sugar allows us to track the molecules but doesn't disturb the chemistry in any way (being radioactive doesn't change the key chemical properties of the molecules relevant to the dynamic equilibrium but does allow us to label the molecules that start in the solid state.

This trick means we can now tell whether there is any exchange between the solid and the liquid even when the solution is saturated. The fact that exchange does happen and radioactive sugar appears in solution means we know the equilibrium is dynamic: this cannot possibly happen if the equilibrium between a solid and a saturated solution is static.

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If the system you describe is in static equilibrium, the solution is saturated and by definition would not accept any more sugar into solution. Since the sugar you're adding is radioactive, this means that you'd observe radioactivity in the solid sugar only.

You don't provide any mechanism by which adding more sugar would disturb this equilibrium. It is true that radioactive sugar will have slightly different properties than normal sugar, but in this case the difference would be imperceptible.

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  • $\begingroup$ At equilibrium, we know that the concentration of the substances involved remain a constant. By adding radioactive sugar, we increase the concentration of sugar. Isn't this disturbing the equilibrium? $\endgroup$ – DoubtExpert May 1 '17 at 14:34
  • $\begingroup$ OK, now I see your confusion. Saturated, by definition, means that the sugar has reached its maximum possible concentration. So you aren't increasing the concentration of the sugar in solution, you're simply increasing the mass of the precipitate. $\endgroup$ – oldchemist May 2 '17 at 1:17
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    $\begingroup$ By adding radioactive sugar, aren't we increasing the mass of the precipitate and disturbing the original mass that was in equilibrium with the dissolved sugar? $\endgroup$ – DoubtExpert May 7 '17 at 16:46
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    $\begingroup$ Adding solid sugar does not disrupt the concentration of solid sugar. The concentration of solid sugar (mass or moles per volume) is a constant - the density of the sugar. Having 200 grams or a 1,000,000 grams does not change the concentration of the solid. $\endgroup$ – Ben Norris Aug 21 '18 at 12:03

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